Question: Evaluate the double integral. $ \int_0^1 \int_{-y^2}^{0} 3x + 2y \, dx \, dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{5}$ (Choice B) B $\dfrac{1}{7}$ (Choice C) C $\dfrac{2}{7}$ (Choice D) D $\dfrac{2}{5}$
Explanation: First, we evaluate the inner integral. We can substitute in the $-y^2$ at the end as if it were a numerical bound. $\begin{aligned} \int_0^1 \int_{-y^2}^{0} 3x + 2y \, dx \, dy &= \int_0^1 \left[ \dfrac{3x^2}{2} + 2xy \right]_{-y^2}^0 dy \\ \\ &= \int_0^1 -\left( \dfrac{3y^4}{2} - 2y^3 \right) \, dy \\ \\ &= \int_0^1 2y^3 - \dfrac{3y^4}{2} \, dy \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_0^1 2y^3 - \dfrac{3y^4}{2} \, dy &= \left[ \dfrac{y^4}{2} - \dfrac{3y^5}{10} \right]_0^1 \\ \\ &= \dfrac{1}{2} - \dfrac{3}{10} \\ \\ &= \dfrac{1}{5} \end{aligned}$ The answer: $ \int_0^1 \int_{-y^2}^{0} 3x + 2y \, dx \, dy = \dfrac{1}{5}$